a) \(x\)=1 \(y\)= 12

b)\(x\)=4 \(y\)= 14

hoặc \(x\)= 6 \(y \)=21

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umk

\(\frac{27}{4}=\frac{-x}{3}=>x=-\frac{81}{4}\notinℤ\)

\(^{y^2=\frac{4}{9}=\left(\frac{2}{3}\right)^2=>y=\pm\frac{2}{3}\notinℤ}\)

\(\frac{27}{4}=\frac{\left(z+3\right)}{-4}=\left(z+3\right)=-27=\left(-3\right)^3=>z+3=-3=>z=-6\)

\(+)|t|-2=-54=>|t|=-52\)(vô lí)

\(+)|t|-2=54=>|t|=56=>t=\pm56\)

a) x=\(\dfrac{5.6}{-10}=-3.\)

b) y=\(\dfrac{3.77}{-33}=-7.\)

x=-3

y=-7

Giải :

\(\dfrac{x-3}{y-2}=\dfrac{3}{2}\) nên 2(x-3) = 3(y-2)

Do đó : 2x - 6 = 3y - 6 nên 2x = 3y

\(\Rightarrow\) 2x - 2y = y hay 2(x-y) = y

Nên 2.4 = y

Vậy : \(y=8;x=\dfrac{3y}{2}=\dfrac{3.8}{2}=12\)

\(\dfrac{x-3}{y-2}=\dfrac{3}{2}\)

\(\Rightarrow\left(x-3\right)\cdot2=3\cdot\left(y-2\right)\)

\(\Rightarrow2x-6=3y-6\)

\(\Rightarrow2x=3y\)

\(\Rightarrow\dfrac{x}{y}=\dfrac{3}{2}\)

mà x - y = 4

\(\Rightarrow\left\{{}\begin{matrix}x=4:\left(3-2\right)\cdot3=12\\y=4:\left(3-2\right)\cdot2=8\end{matrix}\right.\)

a) x.21=6.7

x.21=42

x=42:21

x = 2

b) y . 20 = -5.28

y.20 = -140

y = (-140) : 20

y = -7

a)=>x*21=7*6

=>x*21=42

=>x=42/21

x=2

b)=>y*20=(-5)*28

=>y*20=-140

=>y=-140/20

y=-7

\(\dfrac{-2}{x}=\dfrac{y}{3}\)\(\Rightarrow\left(-2\right).3=x.y\:\Leftrightarrow\:x.y=-6\)

Ta có các cặp số (x;y): \(\left(x=-1;\:y=6\right);\:\left(x=1;\:y=-6\right);\:\left(x=-3;\:y=2\right);\:\left(x=3;\:y=-2\right)\)

Vì \(x< 0< y\) nên có các cặp số thoả mãn: \(\left(x=-1;\: y=6\right);\:\left(x=-3;\: y=2\right)\)

Vậy: \(x=-1;\: y=6\) và \(x=-3\: ;\: y=2\: \)

Bài 1: 

a: \(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2.15+3.75=\dfrac{8}{5}\)

=>x+4/15=8/5 hoặc x+4/15=-8/5

=>x=4/3 hoặc x=-28/15

b: \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{3}x=-\dfrac{1}{6}\\\dfrac{5}{3}x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{6}:\dfrac{5}{3}=\dfrac{-3}{30}=\dfrac{-1}{10}\\x=\dfrac{1}{10}\end{matrix}\right.\)

c: \(\Leftrightarrow\left|x-1\right|-1=1\)

=>|x-1|=2

=>x-1=2 hoặc x-1=-2

=>x=3 hoặc x=-1

Bài 2: 

b: \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Leftrightarrow x=y=-\dfrac{9}{25}\)

Bài 3: 

a: \(A=\left|x+\dfrac{15}{19}\right|-1>=-1\)

Dấu '=' xảy ra khi x=-15/19

b: \(\left|x-\dfrac{4}{7}\right|+\dfrac{1}{2}>=\dfrac{1}{2}\)

Dấu '=' xảy ra khi x=4/7

3. Gọi d là ƯCLN(2n + 3, 4n + 8), d ∈ N*

\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\4n+8⋮d\end{cases}\Rightarrow\hept{\begin{cases}2\left(2n+3\right)⋮d\\4n+8⋮d\end{cases}\Rightarrow}\hept{\begin{cases}4n+6⋮d\\4n+8⋮d\end{cases}}}\)

\(\Rightarrow\left(4n+8\right)-\left(4n+6\right)⋮d\)

\(\Rightarrow2⋮d\)

\(\Rightarrow d\in\left\{1;2\right\}\)

Mà 2n + 3 không chia hết cho 2

\(\Rightarrow d=1\)

\(\RightarrowƯCLN\left(2n+3,4n+8\right)=1\)

\(\Rightarrow\frac{2n+3}{4n+8}\) là phân số tối giản.